Let F be our finite division ring. Let k be its centre.
Note that k is a finite field, as such is has
q elements, for some power of a prime number q.
Consider F as a vector space over k, say it has dimensionn. It follows that F has qn elements.
Consider now the multiplicative groupU(F) of units in F.
Because F is a division ring, U(F) has qn-1
elements.

Next we consider the conjugacy classes of the elements of U(F).
An element of U(F) in the centre has a singleton conjugacy class.
So consider a in U(F) that is not central. The centralizerC of a in F is a subring
of F and obviously it contains k. Further, since a
is not central, C is not equal to F. It follows from the
argument in the first paragraph that C has qd
elements. Because C is itself a division ring
we have that qd-1=|U(C)| is a divisor
of qn-1. It follows quickly that
ddividesn.

Now C(a) the centralizer of a in U(F) consists of the nonzero
elements in C (and so has qd elements) and we
know that the conjugacy class of a in U(F) has
|U(F)|/|C(a)| elements. Since the conjugacy classes of U(F)
give a partition of U(F) we get a formula

qn-1 = q-1 + Sumd (qn-1)/(qd-1) (*)
where the sum is over various proper divisors d of n
and it could be that some ds occur more than once.
(Note that the q-1 comes from the central elements in U(F).)

Now we will make use of cycn(x) the
nth cyclotomic polynomial. It is a factor of any
(xn-1)/(xd-1) for a proper divisor d
of n. So if we evaluate at q we get that
cycn(q) divides each
(qn-1)/(qd-1). For the same reason it also
divides qn-1. So examining (*) we have that
cycn(q) | q-1.

Suppose that n>1. We are going to obtain a contradiction from this.
This will show that n=1 and so F=k is a field.
Since n>1 then |q-e| > q-1 (draw a picture to see this)
and so we see that |cycn(q)| cannot possibly divide
q-1. This contradiction establishes the proof of the theorem.